hdoj1061 Rightmost Digit（快速幂+简单思路）

Description

Given a positive integer N, youshould output the most right digit of N^N. 

Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output
For each test case, you should output the rightmost digit of N^N.

Sample Input
2
3
4

Sample Output
7
6
Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7. In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

题意：求n的n次方的最后一位数。

思路：快速幂，降幂计算不会超时。

代码：

#include<cstdio>__int64 quickpow(__int64 n,__int64 m,__int64 mod){__int64 ans=1,base=n;while(m){if(m&1){ans=(base*ans)%mod;}base=(base*base)%mod;m/=2;}return ans;}int main(){int t;scanf("%d",&t);while(t--){__int64 n;scanf("%I64d",&n);printf("%I64d\n",quickpow(n,n,10));}return 0;}

还有一种方法就是用字符串。记录n的最后一位数，0～9的n次方尾数至多有四种情况。

0.1.5.6.9的n次方还是等于其本身

2=[2,4,8,6],3=[3,9,7,1],4=[4,6],7=[7,9,3,1],8=[8,4,2,6].

所以可以用n对4取余是几就是第几个数。可以打表来做也可以直接写

打表的话char a[4][10]={{0,1,2,3,4,5,6,7,8,9},{0,1,4,9,6,5,6,9,4,1},{0,1,8,7,4,5,6,3,2,9},{0,1,6,1,5,6,1,6,1}};

对四取余的m后直接输出a[m][n].

代码：

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;int main(){int t;scanf("%d",&t);while(t--){int a;scanf("%d",&a);int k=a%10;if(k==0||k==1||k==5||k==6||k==9){printf("%d\n",k);}else{if(k==2){if(a%4==0)printf("6\n");elseprintf("4\n");}if(k==3){if(a%4==1)printf("3\n");elseprintf("7\n");}if(k==4)printf("6\n");if(k==7){if(a%4==1)printf("7\n");elseprintf("3\n");}if(k==8){if(a%4==0)printf("6\n");elseprintf("4\n");}}}return 0;}